How do you find the vertical, horizontal and slant asymptotes of: #f(x)= (x)/(4x^2+7x+-2)#?

Question

Liam Brennan · Accepted Answer

For + sign equation - Horizontal:#larr y = 0rarr#; and vertical : #uarr x =-1.39 and x = -0.34 darr#. For #-# sign- Horizontal: larr y=0rarr#; and vertical: #uarr x= -2 and x=1/4 darr#.

The zeros of the denominator #4x^2+7x+_2# are #-1.39 and -0.34# for + sign #and -2 and 1/4# for - sign. For the straight lines x = these values, #y to +-oo#. Also, as #x to +-oo, y to 0. graph{y(4x^2+7x+2)-x=0 [-20, 20, -10, 10]} graph{y(4x^2+7x-2)-x=0 [-10, 10, -5, 5]}

Cora Alexander · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for x. For this function, the vertical asymptotes occur where the denominator, $4x^2 + 7x - 2$, equals zero. Use factoring or the quadratic formula to find the roots.

To find horizontal asymptotes:
1. If the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes.
2. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0.
3. If the degrees are equal, divide the leading coefficient of the numerator by the leading coefficient of the denominator to find the horizontal asymptote.

To find the slant asymptote:
1. If the degree of the numerator is one greater than the degree of the denominator, perform polynomial long division to find the quotient.
2. The quotient obtained represents the equation of the slant asymptote.

In your case, the vertical asymptotes can be found by setting the denominator $4x^2 + 7x - 2$ equal to zero and solving for x. The horizontal asymptote can be found by comparing the degrees of the numerator and denominator. The slant asymptote can be found if the degree of the numerator is one greater than the degree of the denominator by performing polynomial long division.